Question Number 219864 by Nicholas666 last updated on 02/May/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}^{\:{n}+\mathrm{1}} }{{x}+\mathrm{1}}\:{dx}\:<\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$ \\ $$
Answered by MrGaster last updated on 03/May/25
![(x^(n−1) /(x+1))≤(x^(n+1) /(2(√x)))=(x^(n+0.5) /2) ∫_0 ^1 (x^(n+1) /(x+1))dx≤(1/2)∫_0 ^1 x^(n+0.5) dx=(1/(2(n+1.5))) ∵(1/(2(n+1.5)))< (1/(2(n+1))) [Q.E.D] (2):∫_0 ^1 (x^( n+1) /(x+1))dx≤(1/2)∫_0 ^1 x^(n+(1/2)) dx ∫_0 ^1 x^(n+(1/2)) dx=(1/(n+(3/2)))=(2/(2n+3)) (1/2)∙(2/(2n+3))=(1/(2n+3)) (1/(2n+3))<(1/(2(n+1))) ∴ ∫_( 0) ^( 1) (x^( n+1) /(x+1)) dx < (1/(2(n+1))) [Q.E.D]](https://www.tinkutara.com/question/Q219888.png)
$$\frac{{x}^{{n}−\mathrm{1}} }{{x}+\mathrm{1}}\leq\frac{{x}^{{n}+\mathrm{1}} }{\mathrm{2}\sqrt{{x}}}=\frac{{x}^{{n}+\mathrm{0}.\mathrm{5}} }{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{{x}+\mathrm{1}}{dx}\leq\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{0}.\mathrm{5}} {dx}=\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}.\mathrm{5}\right)} \\ $$$$\because\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}.\mathrm{5}\right)}<\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$$$\left(\mathrm{2}\right):\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\:{n}+\mathrm{1}} }{{x}+\mathrm{1}}{dx}\leq\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\frac{\mathrm{1}}{\mathrm{2}}} {dx}=\frac{\mathrm{1}}{{n}+\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{2}}{\mathrm{2}{n}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{3}}<\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\therefore\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}^{\:{n}+\mathrm{1}} }{{x}+\mathrm{1}}\:{dx}\:<\:\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$
Commented by Nicholas666 last updated on 03/May/25
$${thanks} \\ $$